Proving $\frac1{\sec\phi-\tan\phi} - \frac1{\sec\phi+\tan\phi} =2\tan\phi$

Question

Prove $$\frac1{\sec\phi-\tan\phi} - \frac1{\sec\phi+\tan\phi} =2\tan\phi$$

I have managed to simplify the lefthand side of the equation to $2\sin\phi /\cos^2\phi$ (maybe incorrect) but cannot seem to finish it off.

Answer 1

$$\frac{1}{\sec(\phi)-\tan(\phi)}- \frac{1}{\sec(\phi)+\tan(\phi)}=\frac{\cos\phi}{1-\sin\phi}-\frac{\cos\phi}{1+\sin\phi}=\cos\phi\,\frac{2\sin\phi}{\underbrace{1-\sin^2\phi}_{\cos^2\phi}}=2\,\frac{\sin\phi}{\cos\phi}.$$

Answer 2

HINT:

Obtain a common denominator to arrive at

$$\frac{1}{\sec(\phi)-\tan(\phi)}- \frac{1}{\sec(\phi)+\tan(\phi)}=\frac{2\tan(\phi)}{\sec^2(\phi)-\tan^2(\phi)}$$

Can you finish?

Answer 3

$$\frac1{\sec\theta-\tan\theta}-\frac1{\sec\theta+\tan\theta}=\frac{\sec\theta+\tan\theta-(\sec\theta-\tan\theta)}{\sec^2\theta-\tan^2\theta}=\frac{2\tan\theta}1$$Note the identity $\sec^2\theta=\tan^2\theta+1$

Answer 4

As $(\sec t+\tan t)(\sec t-\tan t)=1,$

for $\sec t\pm\tan t\ne0\iff\sin t\ne\pm1\iff\cos t\ne0$

$$\dfrac1{\sec t\pm\tan t}=\sec t\mp\tan t$$